Question

question write the expression below as a single logarithm in simplest form. \\(\log_{9} 9 + \log_{9} 9\\) answer attempt 1 out of 100

Explanation:

Step1: Recall log addition rule

The logarithmic identity for addition is $\log_b x + \log_b y = \log_b (xy)$. Here, $b = 9$, $x = 9$, and $y = 9$.

Step2: Apply the rule

Substitute into the formula: $\log_9 9 + \log_9 9 = \log_9 (9\times9) = \log_9 81$.

Step3: Simplify $\log_9 81$

Since $9^2 = 81$, by the definition of logarithm $\log_b a = c$ means $b^c = a$, so $\log_9 81 = 2$. But we can also keep it as $\log_9 81$ or recognize the simplification from the addition. Wait, actually, first step: using $\log_b x + \log_b y = \log_b(xy)$, so $\log_9 9 + \log_9 9 = \log_9 (9\times9)=\log_9 81$. Alternatively, since $\log_9 9 = 1$ (because $9^1 = 9$), then $1 + 1 = 2$, but the problem says "as a single logarithm", so we should present it as $\log_9 81$ or simplify further. Wait, $81 = 9^2$, so $\log_9 9^2 = 2$ by the power rule $\log_b b^k = k$. But the question is to write as a single logarithm. So first, apply the addition rule: $\log_9 9 + \log_9 9 = \log_9 (9\times9) = \log_9 81$.

Answer:

$\log_9 81$ (or $2$, but as a single logarithm, $\log_9 81$ is correct. However, since $81 = 9^2$, $\log_9 9^2 = 2$ by power rule, but the single logarithm form is $\log_9 81$)